Answer to Question #238326 in Mechanical Engineering for Richard

Question #238326
Question 2
A flat belt is required to transmit 22 kW from a 250 mm diameter pulley running at 1450
rpm to a 355 mm diameter pulley. The proposed belt is 3.5 mm thick.
The other system specifications are as follows:
 Coefficient of friction 0.7,
 Density of the belt is 1100 kg/m3
 Maximum permissible stress is 7 MPa.
 The distance between the shafts is 1.8 m.
2.1 Calculate the width required. [15]
1
Expert's answer
2021-09-17T02:35:58-0400

Maximum tension T,

P = \frac{2\pi N T}{1000}


T =\frac{22000\times1000}{2\pi\times1450} = 2414.76 N


Width of belt b,

b = \frac{T}{\sigma\times t}= \frac{2414.76}{7\times 3.5}=98.56 mm 






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